functional analysis - Convergence in a norm vs convergence in a strong sense

Let $X = \text{BUC}(\mathbb{R})$ with supremum norm, where BUC is space of functions which are bounded and uniformly continuous.
Let's define an operator:
$$T_n: X \to X$$

$$T_n = f\bigg(x + \frac{1}{n} \bigg).$$
We are to prove that:
$$\lim_{n \to \infty} ||T_nf - f||_{\infty} = 0 \tag{1}$$
but
$$\lim_{n \to \infty} ||T_{n} - I||_{op} \neq 0 \tag{2},$$
where $||T||_{op}$ is the operator norm.

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My solution

Firstly let's prove $(1)$

Because $f \in X$ we can fix $\epsilon >0$ and find $\delta > 0$ such that:
$$\forall_{x,y \in \mathbb{R}} ||x-y||_{\infty} < \delta \implies ||f(x) - f(y)||_{\infty} < \epsilon.$$
Of course $$\forall{\delta>0} \exists_{N \in \mathbb{N}} \forall_{n>N} \bigg| \bigg|\frac{1}{n} \bigg| \bigg| < \delta.$$
Thus
$$\lim_{n \to \infty} \bigg| \bigg| f \bigg(x + \frac{1}{n} \bigg) - f(x) \bigg| \bigg|_{\infty} = \lim_{n \to \infty} \sup_{x \in \mathbb{R}} \bigg|f \bigg(x + \frac{1}{n} \bigg) - f(x) \bigg| = 0.$$

Now let's think about $(2)$

Let's define a sequence of functions: $g_n(x) = \sin(n \pi x)$.
It's obvious that $\forall_{n \in \mathbb{N}} g_n \in X$. Moreover $\forall_{n \in \mathbb{N}} ||g_n||_{\infty} = 1.$
$$||T_n - I||_{op} = \sup_{||f||_{\infty} = 1} ||T_nf - f||_{\infty} \tag{3}.$$
Because we do have the supremum norm over all $f \in X: ||f||_{\infty} = 1$ thus $(3)$ has to be equal or bigger than the following one:
$$||T_ng_n - g_n||_{\infty} = \sup_{x \in \mathbb{R}} |-2 \sin(n \pi x)| = 2.$$
In conclusion
$$\lim_{n \to \infty} ||T_n - I||_{op} \ge 2 \not\to 0.$$
I wonder if my attempts are correct?
If they are I would like to ask why the sequence $g_n$ works in $(2)$ but doesn't in $(1)$? I know that the two norms are something different but honestly I can't see any difference in the calculations. I mean that putting $g_n$ into $(1)$ would lead to something very similar. Where's the difference?

Answer

If by $BUC(\mathbb R)$ you mean the space of bounded and uniformly continuous functions $f:\mathbb R\to\mathbb R$ (or $\mathbb R\to\mathbb C$, whichever you prefer), then the proofs are correct.

I think your confusion comes from the distinction between strong convergence and norm convergence of operators. Think of strong convergence of operators as similar to pointwise convergence of functions from introductory real analysis, and norm convergence of operators as similar to uniform convergence of functions. In pointwise convergence (resp. strong convergence), rates of convergence can vary wildly from point to point (resp. function to function), while in uniform convergence (resp. norm convergence), points (resp. functions) must tend to the limit in a uniform manner.